Eight fair coins tossed simultaneously: What is the probability of getting at least 6 heads (i.e., 6, 7, or 8 heads)?

Introduction / Context:“At least k heads” suggests summing binomial probabilities. For 8 fair coins, we sum for 6, 7, and 8 heads. Each outcome has probability (1/2)^8.

Given Data / Assumptions:n = 8, p(head) = 1/2, independence across coins.

Concept / Approach:Use binomial coefficients C(8,6), C(8,7), C(8,8) and divide by 2^8 = 256.

Step-by-Step Solution:

Verification / Alternative check:By symmetry, P(X ≥ 6) = P(X ≤ 2); indeed C(8,0)+C(8,1)+C(8,2) = 1+8+28 = 37 ⇒ same result.

Why Other Options Are Wrong:Values like 25/57 or 1/13 are unrelated to 2^8 denominators and correct counts.

Common Pitfalls:Forgetting to include all three counts (6,7,8) or miscomputing combinations.

Final Answer:37/256